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karena ρ air = 1 gram /ml, maka massa pelarut air 990 ml adalah 990 gram
mol air = 990/18 = 55
fraksi mol zat terlarut Xt = mol X/ [mol X + mol air]
Xt = (120/Mr) / [(120/Mr) + 55]
tekanan uap jenuh air p₀ = 18 mmHg
besar penurunan tekanan uap adalah Δp = p₀ - p = 18 - 17,37 = 0,63
penurunan tekanan uap larutan Δp = p₀ . Xt
0,63 = (42)x(120/Mr) [(120/Mr) + 55]
0,63. [(120/Mr) + 55] = 42x(120/Mr)
75,6/Mr + 34,65 = 5040 Mr
diperoleh nilai Mr = 49,649
___selesai___